Prove that $a\in A\ \text{and}\ r\in R \implies ar\in A.$

Question

Let $R$ be a commutative ring and $a\in R$. Define $A=\{na:n\in \mathbb{Z^+}\}+aR$. Now I want to prove that $x\in A\ \text{and}\ r\in R \implies xr\in A.$

Maybe I missed a very simple point or overthought things here. I'm not sure. After trying for hours I came up with the following idea. I always seemed to need $R$ to have an identity element. So the ring $R$ is embedded in the ring $R\times\mathbb{Z}$ with identity $(0,1)$ as in this. Now let $x\in A\times\{0\}$. Then $x=(na,0)+(ar,0)$ for some $n\in\mathbb{Z^+}, r\in R$. Let $s=(b,0)\in R\times\{0\}$. Then \begin{align} xs &=((na,0)+(ar,0))(b,0)\\ &=(na,0)(b,0)+(ar,0)(b,0)\\ &=(nab,0)+(arb,0)\\ &=(a,0)+(nab,0)+(arb,0)-(a,0)\\ &=(a,0)+(nab+arb-a,0)\\ &=1(a,0)+(a,0)((nb,0)+(rb,0)-(0,1)). \end{align}

So $\forall x\in A\times\{0\}:\forall s\in R\times\{0\}: xs\in A\times \{0\}$.

I think it can now be inferred that $a\in A\ \text{and}\ r\in R \implies ar\in A.$ But I am not sure how and I don't know if it is simply impossible. Is it really impossible? Could someone please help? Thanks.

Added later: Let $x\in A, r\in R$. Then $(x,0)\in A\times \{0\}, (r,0)\in R\times \{0\}$. So $(x,0)(r,0)=(xr,0)\in A\times\{0\}$, whence $xr\in A$. Is this alright?

So when you say $R$ a commutative ring, it might not have an identity element? — Hamed, Jul 21 '17 at 20:33
Yes. The existence of an identity element is not assumed here. — Janitha357, Jul 21 '17 at 20:35

score 5 · Accepted Answer · answered Jul 21 '17 at 20:35

5

This simply can't be true if $\mathbb Z^+$ is supposed to be $\{1,2,3,\ldots\}$. Take the polynomial ring $\mathbb Z[x]$, and consider ideal $R=(x)$ as a rng in this ring.

Using $a=x$, we have that $A=a\mathbb Z^+ + aR$. Everything in $aR$ contains only terms with monomial degree $2$ or more, and everything in $A$ has a nonzero $\alpha x$ term.

Now $x\in A$, (since $x=a+a0\in A$) and $x\in R$, but $x^2\notin A$, because it does not have a nonzero $\alpha x$ component.

On the other hand, if you had used $\mathbb Z^{\geq 0}$, things would be obvious since in that case $aR\subseteq A$. We would have computed that

$(na+as)r=a(nr+sr)=0a+at\in \mathbb Z^{\geq 0}a+aR$.

answered Jul 21 '17 at 20:35

rschwieb

153,510

I agree. But what is the error in my "proof"? – Janitha357 Jul 21 '17 at 20:57
@Janitha357: Your calculation seems to assume that $(a,0)(0,1)=(a,0)$, but that is not how product rings normally work. And even it it was, $s=(nb,0)+(rb,0)−(0,1)$ would not be in $R\times{0}$. – hmakholm left over Monica Jul 21 '17 at 21:47
yes I see it now. – Janitha357 Jul 21 '17 at 21:51
@HenningMakholm There'a no product ring here. The computation $(a,0)(0,1)=(a,0)$ is correct in the extension referenced. But yeah, the other reason hits the nail on the head. – rschwieb Jul 22 '17 at 00:19

Prove that $a\in A\ \text{and}\ r\in R \implies ar\in A.$

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